使用AJAX，MySQL和PHP添加，删除，更新[重复]

Question

这个问题在这里已有答案：

Why shouldn't I use mysql_* functions in PHP? 15个答案

我的应用程序中有一些Ajax和mysql的问题。我的应用程序有点像用户管理器。我有带有ID，USERNAME，PASSWORD，FIRST_NAME，LAST_NAME和EMAIL字段的MySQL数据库。我列出了每页用户分页10条记录。现在我正在尝试为添加新用户，编辑现有用户以及逐个删除用户提供功能。我有下一个代码：

config.php文件

<?php
$mysql_hostname = "localhost"; 
$mysql_user = "root"; 
$mysql_password = "password"; 
$mysql_database = "mybase"; 
$con = mysql_connect($mysql_hostname, $mysql_user, $mysql_password) or die("Opps error! occurred");
mysql_select_db($mysql_database, $con) or die("Opps error! occurred");
?>

这是数据库连接

**

的index.php

**

<?php
include('config.php');
$per_page = 3; 

//getting number of rows and calculating no of pages

$sql = "select count(*) from users";
$result = mysql_query($sql);
$count = mysql_fetch_row($result);
$pages = ceil($count[0]/$per_page);

?>


<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml" xml:lang="en">
<head>
    <title>Users Manager</title>
    <script type="text/javascript" src="http://ajax.googleapis.com/ajax/
libs/jquery/1.3.0/jquery.min.js"></script>
    <script type="text/javascript" src="pagination.js"></script>    
<style>
body { margin: 0; padding: 0; font-family:Verdana; font-size:15px }
a
{
text-decoration:none;
color:#B2b2b2;

}

a:hover
{

color:#DF3D82;
text-decoration:underline;

}
#loading { 
width: 100%; 
position: absolute;
}

#pagination
{
text-align:center;
margin-left:120px;

}
li{ 
list-style: none; 
float: left; 
margin-right: 16px; 
padding:5px; 
border:solid 1px #dddddd;
color:#0063DC; 
}
li:hover
{ 
color:#FF0084; 
cursor: pointer; 
}
</style>
</head>
<body>
<div align="center">
   <div style="margin-top:50px;"><b>Title</b>: users Manager</div>
 <div id="content" ></div>
    <table width="800px">
    <tr><Td>
            <ul id="pagination">
                <?php
                //Show page links
                for($i=1; $i<=$pages; $i++)
                {
                    echo '<li id="'.$i.'">'.$i.'</li>';
                }
                ?>
           </ul>    
        </Td>
    </tr></table>
<div id="loading" ></div>
    </div> 
</body>
</html>

**

data.php

**

    <?php
include('config.php');
$per_page = 10; 
if($_GET)
{
    $page=$_GET['page'];
}

//getting table contents
$start = ($page-1)*$per_page;
$sql = "select * from users order by id limit $start,$per_page";
$rsd = mysql_query($sql);

if(isset($_POST['buttonsave'])){
    $query_sql = "INSERT INTO users (username,firstname,lastname,email) VALUES ('{$_POST[username]}','{$_POST[firstname]}','{$_POST[lastname]}','{$_POST[email]}')";
    $result = mysql_query($query_sql);
    if($result){
        echo "Successful insert";

    }
}
?>
<script type="text/javascript" src="http://ajax.googleapis.com/ajax/
libs/jquery/1.3.0/jquery.min.js"></script>


<table id="tbl">
        <th>User Name:<input type="text" id="username" name="username" placeholder="User"></th>
        <th>First Name:<input type="text" id="firstname" name="firstname" placeholder="First Name"></th>
        <th>Last Name:<input type="text" id="lastname" name="lastname" placeholder="Last Name"></th>
        <th>E-mail:<input type="email" id="email" name="email" placeholder="Email"></th>        
</table>
<input type="button" value="Insert" id="save">

<script type="text/javascript">
$(function(){
    $("#save").click(function(){
        var uname = $("#username").val();
    var fname = $("#firstname").val();
    var lname = $("#lastname").val();
    var email = $("#email").val();
    $.ajax({
        url: "data.php",
        type: "POST",
        async: true,
        data:   {
                    buttonsave: 1,
                    username: uname,
                    firstname: fname,
                    lastname: lname,
                    email: email
                },
        success: function(result){
            alert("OK! Good!");
        }
    });
    });
});
</script>

<table id="tbl">
        <th><b>Id</b></th>
        <th><b>User Name</b></th>
        <th><b>First Name</b></th>
        <th><b>Last Name</b></th>
        <th><b>E-mail</b></th>
        <?php
        while($row = mysql_fetch_array($rsd))
        {
            $id    = $row['id'];
            $uname = $row['username'];
            $fname    = $row['first_name'];
            $lname    = $row['last_name'];
            $email    = $row['email'];
        ?>
        <tr>
        <td><?php echo $id; ?></td>
        <td><?php echo $uname; ?></td>
        <td><?php echo $fname; ?></td>
        <td><?php echo $lname; ?></td>
        <td><?php echo $email; ?></td>
        </tr>
        <?php
        } //End while
        ?>
</table>

<style>
#tbl
{
width:800px;
border:1px solid #2E8AE6;
margin-top:50px;

}

#tbl tr:nth-child(odd) {
  background: #C2E0FF;

}

#tbl td{
border:1px solid #2E8AE6;
padding: 5px;
}

#tbl th
{
  background: #2E8AE6;
border:1px solid #2E8AE6;
padding: 5px;
}
</style>

和我的jQuery文件：**

pagination.js

**

$(document).ready(function(){

    //Loading Image Display
    function Display_Load()
    {
        $("#loading").fadeIn(100);
        $("#loading").html("<img src='loading.gif' />");
    }
    //Hide Loading Image
    function Hide_Load()
    {
        $("#loading").fadeOut('slow');
    };


   //Default Starting Page Results

    $("#pagination li:first").css({'color' : '#FF0084','border' : 'none'});
    $("#content").load("data.php?page=1", Hide_Load());

    //Pagination Click
    $("#pagination li").click(function(){
        Display_Load();

        //CSS Styles
        $("#pagination li")
        .css({'border' : 'solid #dddddd 1px'})
        .css({'color' : '#0063DC'});

        $(this)
        .css({'color' : '#FF0084'})
        .css({'border' : 'none'});

        //Loading Data
        var pageNum = this.id;
        $("#content").load("data.php?page=" + pageNum, Hide_Load());
    });
});

所以问题出在我的data.php中。我试图使用ajax插入数据，但它不起作用！它不会向数据库插入任何内容。但它提醒我

alert("OK! Good!");

我无法理解那是错的。请帮忙。